# Equivalent resistance in multiloop

1. Homework Statement 2. Homework Equations
Series: R(eq) = R(1) + R(2) + . . .

Parallel: $$\frac{1}{R(eq)}$$ = $$\frac{1}{R(1)}$$ + $$\frac{1}{R(2)}$$ + . . .

3. The Attempt at a Solution

I'm confident w/ (b) i think, but i just want to know if i'm going about these correctly.

for (a), since nothing is in the center wire, that can be treated as a node. If that's the case, the top 2 and the bottom 2 can be treated like they're in parallel respectively, then added like they are in series. R(eq) = R(1)

for (b), the two sides can be treated like they are in series, then those are added in parallel. R(eq) = $$\frac{R(1)}{2}$$

for (c), I'm labeling directions of current, but i'm not sure which add in what way. Any help would be great.

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Double check your math for (b).
Don't make (c) harder than it is (look at the hint).

To make things easier, the 2 junctions in the middle due to the wire in the middle will be labelled as points c and d. For each of the 3 circuits there is a junction next to A and B, in which the current splits along 2 routes. In both routes, whether it goes from A to B or vice versa, they go through a resistor R1 before making it to points c and d. Since they undergo the same resistance, the voltage drop for each route will be the same, so in that case, the electric potential at c would be the same at d...therefore, the potential difference between c and d is 0V (meaning there is no current running along that wire). So in circuits (a) and (c), there is no current in the wire in the middle and it can be excluded.

To make things easier, the 2 junctions in the middle due to the wire in the middle will be labelled as points c and d. For each of the 3 circuits there is a junction next to A and B, in which the current splits along 2 routes. In both routes, whether it goes from A to B or vice versa, they go through a resistor R1 before making it to points c and d. Since they undergo the same resistance, the voltage drop for each route will be the same, so in that case, the electric potential at c would be the same at d...therefore, the potential difference between c and d is 0V (meaning there is no current running along that wire). So in circuits (a) and (c), there is no current in the wire in the middle and it can be excluded.

Ok, the math was wrong in (b), R(eq) in (b) = R(1)

Now in (c), i totally understand what you're saying. So if that's the case, the equivalent resistance through all 3 circuits is just R(1).

What about a situation if there were a R(1) and a R(2) on opposite sides in (c)?

bump...^^